# Proving that $\sum_{k=0}^{\infty}\frac{1}{(k+1)(2k+1)(4k+1)}=\frac{\pi}{3}$

The following problem(p.668, 7) is from Integrals and Series [ Интегралы и ряды, А.П. Прудников, Ю.А. Брычков, О.И. Маричев.] states that

$$\sum_{k=0}^{\infty}\frac{1}{(k+1)(2k+1)(4k+1)}=\frac{\pi}{3}$$

How one can show that?

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Hint: re-write as $\frac{1}{3}\int_0^1 \sum x^k dx-2\int_0^1 \sum x^{2k}dx+\frac{8}{3} \int_0^1 \sum x^{4k}\,dx$. Can you see what to do next? – L. F. Dec 10 '13 at 12:05
See the answers on this related post. – Lucian Dec 10 '13 at 12:26
I'd say the title would be more commonly translated as "Integrals and Series" – DonAntonio Dec 10 '13 at 13:55
@DonAntonio: You are right! :) – Salech Alhasov Dec 10 '13 at 13:57